1018. Public Bike Management (30) 最短路问题

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先马一下做法，这道题就是先找最短路，然后求出所有的最短路中，send 最少的，然后再求出back最少的。

我在网上也找代码看了看，各位大神好像都喜欢把解法叫成 dijkstra + dfs

首先dfs是什么深度优先搜索。这道题是什么？回溯啊，我不知道这是误区还是约定俗成，所以我还是坚持我的看法 dijkstra + 回溯或者 bfs找最短路 + 回溯这样起码不会看到就

引起误会。虽然dfs和回溯都是用递归，但是思想上是不同的。个人看法各位大神勿喷。

我感觉按照pat的尿性只要是涉及到算法的问题，首先想回溯就可以了。

今天不想写了。明天写写贴个代码。

#include<cstdio>#include<cstring>#include<cmath>#include<string>#include<vector>#include<map>#include<set>#include<queue>#include<algorithm>#include<iostream>#include<string>using namespace std;#define INF 0x6fffffffint cmax, n, sp, m;int num[510];int e[510][510];int dis[510],book[510];int shortlen,vis[510],path[510];bool finsh = false;int nowstep,in=INF,out=INF;int rstep, rpath[510];void init() {int i, j;for (i = 0; i <= n; i++){for (j = 0; j <= n; j++){e[i][j] = INF;}}}void dijkstra() {int i, j, u, v;for ( i = 0; i <= n; i++){int min = INF;for ( j = 0; j <= n; j++){if (book[j]==0 && dis[j] < min){min = dis[j];u = j;}}book[u] = 1;for (v = 0; v <= n; v++){if (e[u][v] < INF&&!book[v]){if (dis[v] > dis[u] + e[u][v]){dis[v] = dis[u] + e[u][v];}}}}shortlen = dis[sp];}void cal() {int tin = 99999999, tout = 999999999;int i, add = 0;for (i = 1; i <= nowstep; i++) {add += num[path[i]] - cmax / 2;if (add < tin)tin = add;}if (tin > 0)tin = 0;else tin = -tin;tout = add + tin;if ((tin < in) || ((tin == in) && tout < out)) {in = tin;out = tout;if (in == 0 && tout == 0) finsh = true;rstep = nowstep;for (i = 1; i <= nowstep; i++) {rpath[i] = path[i];}}}void dfs(int st,int first,int len) {if (finsh || len > shortlen) return;if (first == sp&&len == shortlen){nowstep = st;cal();}int j;for (j = 0; j <= n; j++){if (vis[j] == 1 || e[first][j] == INF){continue;}vis[j] = 1;path[st+1] = j;dfs(st + 1, j, len + e[first][j]);vis[j] = 0;}}void print() {int i;printf("%d", in);printf(" 0");for (i = 1; i <= rstep; i++) {printf("->%d", rpath[i]);}printf(" %d\n", out);}int main(){cin >> cmax >> n >> sp >> m;for (int i = 1; i <= n; i++) {cin >> num[i];}init();for (int i = 0; i < m; i++){int t1, t2, t3;cin >> t1 >> t2 >> t3;e[t1][t2] = e[t2][t1]=t3;}for (int i = 0; i <= n; i++){dis[i] = e[0][i];book[i] = 0;}book[0] = 1;dijkstra();dfs(0, 0, 0);print();return 0;}

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